Abstract Nonsense

Crushing one theorem at a time

Left Regular Representation (Pt. II)

Point of post: This post a is a continuation of this one.

Left Multiplication Representation

With the above inner product we can now successfully describe the canonical \ast-representation of \mathcal{A}\left(G\right) into \text{End}\left(\mathcal{A}\left(G\right)\right) endowed with the canonical inner product. Namely:

Theorem: Let G be a finite group and \mathcal{A}\left(G\right) be the group algebra on G. Then, if \mathcal{A}\left(G\right) is given the canonical inner product then the map \varrho:\mathcal{A}\left(G\right)\to\text{End}\left(\mathcal{A}\left(G\right)\right) given by \varrho_a(b)=a\ast b is a \ast-representation.

Proof: First of all it’s worth showing that the map \varrho_a:\mathcal{A}\left(G\right)\to\mathcal{A}\left(G\right) really does define a linear transformation. But, this is fairly trivial to check since for any \beta,\gamma\in\mathbb{C} and a,b,c\in\mathcal{A}\left(G\right)


\begin{aligned}\varrho_a\left(\beta b+\gamma c\right) &= a\ast\left(\beta b+\gamma c\right)\\ &= a\ast\left(\beta b\right)+a\ast\left(\gamma c\right)\\ &= \beta\left(a\ast b\right)+\gamma\left(a\ast c\right)\\ &= \beta\varrho_a(b)+\gamma\varrho_a(c)\end{aligned}


It thus remains to show that \varrho is a \ast-representation of \mathcal{A}\left(G\right). To see that \varrho is additive we merely note that for any a,b,c\in\mathcal{A}\left(G\right) we have that

\displaystyle \varrho_{a+b}(c)=\left(a+b\right)\ast c=a\ast c+b\ast c=\varrho_a(c)+\varrho_b(c)


from where it follows from the arbitrariness of c that \varrho_{a+b}=\varrho_a+\varrho_b. Similarly one has that

\displaystyle \varrho_{a\ast b}(c)=(a\ast b)\ast c=a\ast (b\ast c)=\varrho_a(b\ast c)=\varrho_a\varrho_b(c)


and thus by the arbitrariness of c it follows that \varrho_{a\ast b}=\varrho_a\varrho_b. Noticing that evidently for any a\in\mathcal{A}\left(G\right) one has that \varrho_{\delta_e}(a)=\delta_e\ast a=a it’s clear that \varrho_{\delta_e}=\mathbf{1}. Thus, it remains show that that \displaystyle \varrho_{a^\ast}=\varrho_a^\ast. To do this we recall that \varrho_a^\ast is the unique endomorphism on \mathcal{A}\left(G\right) such that

\left\langle \rho_a(b),c\right\rangle=\left\langle b,\rho_a^\ast(c)\right\rangle


for all b,c\in\mathcal{A}\left(G\right).  Thus, it suffices to show that \rho_{a^\ast} satisfies this property. Indeed, let b,c\in\mathcal{A}\left(G\right) and note that


\displaystyle \begin{aligned}\left\langle \rho_a(b),c\right\rangle &= \left\langle a\ast b,c\right\rangle\\ &= \frac{1}{|G|}\sum_{g\in G}(a\ast b)(g)\overline{c(g)}\\ &= \frac{1}{|G|}\sum_{g\in G}\sum_{k\in G}a\left(gk^{-1}\right) b(k)\overline{c(g)}\\ &= \frac{1}{|G|}\sum_{g\in G}\sum_{k\in G}\overline{\overline{a\left(\left(kg^{-1}\right)^{-1}\right)}}b(k)\overline{c(g)}\\ &= \frac{1}{|G|}\sum_{g\in G}\sum_{k\in G}\overline{a^{\ast}\left(kg^{-1}\right)}b(k)\overline{c(g)}\\ &= \frac{1}{|G|}\sum_{k\in G}\left(\sum_{g\in G}\overline{a^\ast(kg^{-1})c(g)}\right)b(k)\\ &= \frac{1}{|G|}\sum_{k\in G}b(k)\overline{\left(a^\ast\ast c\right)(k)}\\ &= \left\langle b,a^\ast \ast c\right\rangle\\ &= \left\langle b,\rho_{a^{\ast}}(c)\right\rangle\end{aligned}


from where by the arbitrariness of a,b\in\mathcal{A}\left(G\right) we are able to conclude, by the aforementioned characterization of the adjoint, that \rho_{a^\ast}=\rho_a^\ast. The theorem then follows. \blacksquare


Left Regular Representation

We have just seen that the map \varrho:\mathcal{A}\left(G\right)\to\text{End}\left(\mathcal{A}\left(G\right)\right) given by \varrho_a(b)=a\ast b is a \ast-representation on \mathscr{A}\left(G\right). But, by previous theorem we know that this induces a representation \rho:G\to\mathcal{U}\left(\mathcal{A}\left(G\right)\right) given by g\mapsto \varrho(\delta_g). At first this may seem like quite an intractable definition since, in practice, \varrho(\delta_g) means nothing. What’s interesting is that \varrho(\delta_g) acts on elements of \mathcal{A}\left(G\right) in a particularly simple way. In particular, \left(\rho_g(a)\right)(h)=a\left(g^{-1}h\right). Indeed:


\displaystyle \begin{aligned}\left(\rho_g(a)\right)(h) &= \left(\varrho_{\delta_g}(a)\right)(h)\\ &= \left(\delta_g\ast a\right)(h)\\ &= \sum_{k\in G}\delta_g\left(hk^{-1}\right)a(k)\\ &= \delta_g\left(h\left(g^{-1}h\right)^{-1})\right)a\left(g^{-1}h\right)\\ &= a\left(g^{-1}h\right)\end{aligned}


This representation is called the left regular representation of G on \mathcal{A}\left(G\right).

Now, while this is an enlightening way to look at the left regular representation there is an admittedly easier method to prove that it’s a representation. Namely, it’s clear that if \rho is the left regular representation then \rho is a homomorphism G\to\text{End}\left(\mathcal{A}\left(G\right)\right) and thus it would suffice to check that \rho_g is unitary for each g\in G. Note though that \rho_g(\delta_h)=\delta_{gh} from where it clearly follows that \rho_g takes the orthonormal basis \{\sqrt{|G|}\delta_h\}_{h\in G} to itself, and thus (by a common characterization of unitarity) each \rho_g is unitary.


We shall see that the left regular representation will play a decisive role in the theory to come.



1.Simon, Barry. Representations of Finite and Compact Groups. Providence, RI: American Mathematical Society, 1996. Print.

2. Serre, Jean Pierre. Linear Representations of Finite Groups. New York: Springer-Verlag, 1977. Print



January 23, 2011 - Posted by | Algebra, Representation Theory | , ,


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